\(^{\left|5x-31\right|.3^2.10\left(-100\right).25}\)= bao nhiu
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10 tháng 7 2023
\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)
\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)
10 tháng 7 2023
a: =>(x+2-3)(x+2+3)=0
=>(x-1)(x+5)=0
=>x=1 hoặc x=-5
b: =>(x-1)^2=25
=>x-1=5 hoặc x-1=-5
=>x=-4 hoặc x=6
c: =>25x^2+10x+1-25x^2+9=30
=>10x+10=30
=>x+1=3
=>x=2
d: =>x^3-1-x(x^2-4)=5
=>x^3-1-x^3+4x=5
=>4x=6
=>x=3/2
24 tháng 8 2021
Ta có: \(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+3\right)^3+\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3\)
\(=x^3+125-x^3-9x^2-27x-27+x^3-8-x^3+3x^2-3x+1\)
\(=-6x^2-30x+91\)
NL
1
15 tháng 12 2023
\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
=>\(x^3-6x^2+12x-8-\left(x^3+125\right)+6x^2=11\)
=>\(x^3+12x-8-x^3-125=11\)
=>12x-133=11
=>12x=144
=>\(x=\dfrac{144}{12}=12\)
JA
2
22 tháng 3 2016
Bài này dễ mà:
A=(100-1^2)(100-2^2)(100-3^2)...(100-10^2)...(100-25^2)
A=(100-1)(100-4)(100-9)...(100-100)...(100-625)
=> A=0
18 tháng 7 2017
\(\left(x-4\right)^3-\left(x-5\right)\left(x^2+5x+25\right)=\left(x+2\right)\left(x^2-2x+4\right)-\left(x+4\right)^3\)
\(\Leftrightarrow x^3-12x^2+48x-64-x^3+125=x^3+8-x^3-12x^2-48x+64\)
\(\Leftrightarrow-12x^2+48x+61=-12x^2-48x+72\)
\(\Leftrightarrow48x+61=-48x-72\)
\(\Leftrightarrow x=\dfrac{-133}{96}\)
NA
1
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 12 2020
ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\x\ge\dfrac{3}{5}\end{matrix}\right.\)
\(\left(x+1\right)\left(45x^2-62x+25\right)=4\sqrt{\left(x+1\right)\left(5x-3\right)\left(5x-3\right)^2}\)
- Với \(x=-1\) là 1 nghiệm
- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT< 0\\VP>0\end{matrix}\right.\) pt vô nghiệm
Với \(x\ge\dfrac{3}{5}\) ta có:
\(45x^3-17x^2-37x+25=4\sqrt{\left(x+1\right)\left(5x-3\right)\left(5x-3\right)^2}\)
\(\Leftrightarrow45x^3-17x^2-37x+25\le2\left[\left(x+1\right)\left(5x-3\right)+\left(5x-3\right)^2\right]\)
\(\Leftrightarrow45x^3-77x^2+19x+13\le0\)
\(\Leftrightarrow\left(x-1\right)^2\left(45x+13\right)\le0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Y
7 tháng 7 2023
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)
\(10,\left(x+3\right)^2-x^2=45\)
\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)
Vậy \(S=\left\{6\right\}\)
\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)
\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)
DN
24 tháng 5 2017
a)
\(A=\left(x-6\right)^2+\left(x+6\right)^2\)
\(A=\left(x^2-2x6+6^2\right)+\left(x^2+2x6+6^2\right)\)
\(A=x^2-2x6+6^2+x^2+2x6+6^2\)
\(A=\left(x^2+x^2\right)+\left(-2x6+2x6\right)+\left(6^2+6^2\right)\)
\(A=2x^2+72\)
b)
\(B=\left(x^2+y^2+3^2+2xy+2x3+2y3\right)-\left(x^2+y^2+9\right)\)
\(B=x^2+y^2+3^3+2xy+2x3+2y3-x^2-y^2-9\)
\(B=\left(x^2-x^2\right)+\left(y^2-y^2\right)+\left(3^2-9\right)+2xy+2x3+2y3\)
\(B=2xy+2x3+2y3\)
Mình phải đi ngủ rồi, có gì mai làm tiếp nha 
DN
25 tháng 5 2017
c/
C = (5x - 2) . (5x + 2) - (5x - 1)2
C = [(5x)2 - 22] - [(5x)2 - 2 . 5x1 + 12]
C = (5x)2 - 22 - (5x)2 + 2 . 5x1 - 12
C = [(5x)2 - (5x)2] + (-22 + 2 - 12) + 5x1
C = 5 + 5x1.
\(\left|5x-31\right|.3^2.10\left(-100\right).25\)
\(=\left|5x-31\right|.9.10.\left(-2500\right)\)
\(=\left|5x-31\right|.\left(-225000\right)\)
bài này có biến \(x\) mà cũng đòi ra kết quả hả bạn